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Question

The natural frequency of free torsional vibrations of a shaft with torsional stiffness q and I is the mass moment of inertia of the disc attached to the end of the shaft is

The correct answer is \(\frac{1}{{2\pi }} \times \sqrt {\frac{q}{I}}\)

Understanding Torsional Vibration Frequency of a Shaft

This explanation covers the calculation for the natural frequency of free torsional vibrations of a shaft, considering its torsional stiffness and the mass moment of inertia of an attached disc.

Deriving the Natural Frequency Formula

Torsional vibrations occur when a shaft is twisted and then released, causing it to oscillate back and forth around its equilibrium position. The key parameters influencing this vibration are:

  • Torsional Stiffness (q): This represents the shaft's resistance to twisting. A higher stiffness means more force is required to produce a given angle of twist.
  • Mass Moment of Inertia (I): This measures the resistance of the disc attached to the shaft's end to angular acceleration. It depends on the disc's mass and how it's distributed relative to the axis of rotation.

Calculation of Angular Frequency

For a system exhibiting simple harmonic motion, like torsional vibrations, the angular frequency ($\omega$) is related to the system's properties. The formula for the angular frequency of a torsional pendulum is:

$ \omega = \sqrt{\frac{\text{Restoring Force}}{\text{Inertia}}} $

In the case of torsional vibrations:

  • The "restoring force" analogue is the restoring torque, which is proportional to the angle of twist. The torsional stiffness '$q$' relates torque to twist angle ($\tau = q\theta$).
  • The "inertia" is the mass moment of inertia '$I$' of the disc.

Therefore, the angular frequency ($\omega$) is given by:

$ \omega = \sqrt{\frac{q}{I}} $

Converting Angular Frequency to Natural Frequency

The natural frequency ($f$), measured in Hertz (Hz) or cycles per second, is related to the angular frequency ($\omega$) in radians per second by the following relationship:

$ \omega = 2\pi f $

To find the natural frequency ($f$), we rearrange this formula:

$ f = \frac{\omega}{2\pi} $

Substituting the expression for $\omega$ we found earlier:

$ f = \frac{1}{2\pi} \sqrt{\frac{q}{I}} $

Conclusion

Based on the derivation, the natural frequency of free torsional vibrations for the given shaft system is correctly represented by the formula:

\(\frac{1}{{2\pi }} \times \sqrt {\frac{q}{I}}\)

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Important Questions from Torsional Vibration

  1. Torsional vibrations on a crankshaft is reduced by ______.
  2. A shaft which is 50 mm diameter and 3 metres long is simply supported at the ends and carries three loads of 1000 N, 1500 N and 750 N at 1 m, 2 m and 2.5 m from the left support. The Young's modulus for shaft material is 200 \(\rm \frac{GN}{m^2}\). Determine the frequency of transverse vibration.

  3. Consider a uniform shaft of length L fixed at its upper end and carrying a disc of the moment of inertia I at its lower end. The disc is twisted about the vertical axis and released. 'fa' is the natural frequency of the system when the shaft is assumed as massless, and 'fb' is the natural frequency of the system when the shaft is considered of the same moment of inertia as that of the disc. Find the ratio fa/fb.

  4. A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; if two masses each of 'm' are attached at distance 'L/2' from its centre of both sides, it reduces the oscillation frequency by 10%. The value of the ratio M/m is close to :
  5. A solid steel shaft transmits 40 kW of power at a speed of \(\frac{75}{\pi}\)Hz. The internal torque needed in the shaft is

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